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Which is the limiting reactant in the following reaction given 42.0 g of CO₂ reacts with 99.9 g of KOH?

\[ \text{CO}_2 + 2\text{KOH} \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O} \]

Answer :

Final answer:

To determine the limiting reactant in the given reaction, we compare the number of moles of each reactant to their stoichiometric coefficients. By converting the masses of CO₂ and KOH to moles, we find that there is an excess of KOH and CO₂ is the limiting reactant.

Explanation:

To determine the limiting reactant, we need to compare the number of moles of each reactant to the stoichiometric coefficients in the balanced equation.

First, we convert the mass of CO₂ to moles using its molar mass:

42.0 g CO₂ * (1 mol CO₂ / 44.01 g CO₂) = 0.955 mol CO₂

Next, we convert the mass of KOH to moles using its molar mass:

99.9 g KOH * (1 mol KOH / 56.11 g KOH) = 1.78 mol KOH

According to the balanced equation, the stoichiometric ratio between CO₂ and KOH is 1:2. Therefore, for every 1 mole of CO₂, we need 2 moles of KOH. Comparing the number of moles we calculated, we can see that there is an excess of KOH, making it the excess reactant, and CO₂ is the limiting reactant.

Therefore, CO₂ is the limiting reactant in this reaction.

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